іmрlеmеոt thе fоllоwіոg ехрrеssіоո іո аssеmbly lаոguаgе: Create a DWORD array named ‘z’ of size 3 using DUP operator. Leave the array ‘z’ uninitialized. You can denote the items in the array as [ z0, z1, z2]z0, z1, z2] , where z0z0 is the first item, z1 z1 is the second item, z2z2 is the third item Update each array item using the following expressions.

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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 іmрlеmеոt thе fоllоwіոg ехрrеssіоո іո аssеmbly lаոguаgе:
  • Create a DWORD array named ‘z’ of size 3 using DUP operator. Leave the array ‘z’ uninitialized. You can denote the items in the array as [

    z0, z1, z2]z0, z1, z2]

    , where

    z0z0

    is the first item,

    z1 z1

    is the second item,

    z2z2

    is the third item
  • Update each array item using the following expressions.

  

                                               

z0=x+130z0=x+130

  

z1=y−x+ z0z1=y−x+ z0

  

z2=r+x−z1z2=r+x−z1

  

                                                                   

  • Where x, y, r are 16-bit integer memory variables.  

  • x = 10, y = 15, r = 4

  • Use mov, movzx, movsx, add, sub instructions only.  

  • (hint: Do not alter the value of x, y and r during the computation. Transfer them to appropriate registers to do computation)

  • At the end, open memory window to see the variable z stored in memory (little endian format).

  • Use the debugger to verify your answer.

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